# The x components of the average accelerations?

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1. Oct 26, 2016

### emily081715

1. The problem statement, all variables and given/known data
A(n) 8.5-kg object is sliding across the ice at 2.34 m/s in the positive x direction. An internal explosion occurs, splitting the object into two equal chunks and adding 12 J of kinetic energy to system. The explosive separation takes place over a 0.16-s time interval. Assume that the one of the chunks after explosion moves in the positive x direction. The x component of the average acceleration of this chunk during the explosion is afrontx, the x component fo the average acceleration of the other chunk during the explosion is arearx. What are the x components of the average accelerations of the two chunks during the explosion?

2. Relevant equations
k=1/2mv^2

3. The attempt at a solution
k=1/2(8.5)(2.34)^2
= 23.27J

23.27+12=35.27J
35.27/0.16= 220.4m/s^2
i've never done a problem like this

2. Oct 26, 2016

### Aniruddha@94

What values are required to compute the average acceleration? ( or what's the formula for it).

3. Oct 26, 2016

### emily081715

a=vf-vi/t

4. Oct 26, 2016

### Aniruddha@94

Right. You've been given $t$ and the initial velocity for both the pieces. You need to find their final velocities. There are two unknowns, so we'll need 2 equations to solve them. Can you see what the equations can be?

5. Oct 26, 2016

### emily081715

is one going to be the change in kinetic energy formula, and is the change momentum??

6. Oct 26, 2016

### emily081715

is the initial velocity for both pieces just 2.34?

7. Oct 26, 2016

### Aniruddha@94

Yes. You'll solve them to get the final velocities of both the pieces.
That's right.

8. Oct 26, 2016

### emily081715

for the mass do i use 8.5, or need to divide that by two since its split equally?

9. Oct 26, 2016

### Aniruddha@94

When expressing the final momentum ( or kinetic energy) of the system, you'll have to divide it by two.

10. Oct 26, 2016

### emily081715

but on the initial side, it would just be 8.5(2.34) for momentume, and 1/2(8.5)(2.34)^2 for kinetic energy?

11. Oct 26, 2016

### Aniruddha@94

Yes, that's correct.

12. Oct 26, 2016

### emily081715

i got answers of 20.48 and 25.16. they're wrong

13. Oct 26, 2016

### emily081715

14. Oct 26, 2016

### emily081715

that was my work

15. Oct 26, 2016

### Aniruddha@94

Your application of momentum conservation doesn't look right.
Suppose $v$ to be the initial velocity of the block and $v_1$ and $v_2$ be the final velocities of the two pieces. If the mass of the block is $m$, the conservation principle gives $$p_i = p_f$$
$$mv = \frac {m}{2}v_1 + \frac{m}{2}v_2$$
This is your first equation. Similarly, for kinetic energy you'll get $$\frac{1}{2}mv^2 + 12 = \frac{1}{2}\frac{m}{2}(v_1)^2 +\frac{1}{2}\frac{m}{2}(v_2)^2$$
Solve these two equations for $v_1$and $v_2$. Use them in the formula for average acceleration.

16. Oct 26, 2016

### emily081715

i got answers of 74.56 and -74.31 for acceleration which are wrong

17. Oct 26, 2016

### Aniruddha@94

That doesn't match with my results either.
What values did you get for $v_1$ and $v_2$ ? Show your work.

18. Oct 26, 2016

### emily081715

i go values of 14.27 m/s and -9.59 m/s

19. Oct 27, 2016

### Aniruddha@94

In the third line, you have substituted for $v_2$ in terms of $v_1$. You seem to have made a mistake there.
From the first equation(momentum conservation), we get $2v = v_1 + v_2$. Or $v_2 = 4.68 - v_1$ . Use this in your second equation.

20. Oct 27, 2016

### emily081715

i got accelerations of -117.136 m/s^2 and 117.136m/s^2